Question

John kept $\frac{4}{12}$, $\frac{3}{12}$, and $\frac{2}{12}$ pieces of a puzzle in three bags.
He kept the remaining pieces in the fourth bag.

• A. $\frac{4}{12}$
• B. $\frac{1}{12}$
• C. $\frac{2}{12}$
• D. $\frac{3}{12}$

Answer 1

The correct option is D $\frac{3}{12}$

Calculate the fraction of puzzle pieces to be kept in the fourth bag.

So, let's calculate the total number of pieces in first three bags.

Fraction of puzzle pieces in the first three bags = $\frac{4}{12}+\frac{3}{12}+\frac{2}{12}$ (like fractions)

As the denominators of the fractions are same, the numerators can be added directly and the denominator remains same.

$=(\frac{4+3+2}{12})$

$=\frac{9}{12}$

So, the fraction of pieces in three bags is $\frac{9}{12}$.

To find the fraction of pieces in the fourth bag, subtract $\frac{9}{12}$ from the total number of pieces, i.e., $\frac{12}{12}$

$=\frac{12}{12}-\frac{9}{12}$ (like fractions)

As the denominators of the fractions are same, the numerators can be subtracted directly and the denominator remains same.

$=\frac{12-9}{12}$

$=\frac{3}{12}$

Hence, $\frac{3}{12}$ puzzle pieces are kept in the fourth bag.